Method other than for loop?

2011 6 17
2 个回答
回答已采纳
3 次查看（30 天）

0 个投票

My code is quite long and outputs two values called beam1 and beam2. The values of these variables are dependant on another variable labeled length.
What I want my code to do is output a length value that allows beam1-beam2<0.0001. As of now I am using a for loop to cycle through a range of different length values to output the ones that fit my criteria. However this method takes an extremely long time to run.
Do you have any suggestions that I could use instead of a for loop that might help the program take less time to run?

2 个评论
显示 无 隐藏 无

Sean de Wolski
Sean de Wolski 2011-6-17
Don't name your variable 'length' as that is a very useful MATLAB function that you don't want to have overwritten.
Sean de Wolski
Sean de Wolski 2011-6-17
Other than that we're probably going to have to see the code (at least the relevant portions) to help.

请先登录，再进行评论。

请先登录，再回答此问题。

 采纳的回答

Walter Roberson
Walter Roberson 2011-6-17

0 个投票

You could possibly use something like fmincon to minimize on (beam1-beam2).^2 with a tolerance of 0.0001^2 . Ummm -- can your code come out with beam2 > beam1 ?
Does your computation have a lot of local minima? If so then a more powerful minimizer might be needed.

9 个评论
显示 7更早的评论 隐藏 7更早的评论

Derek
Derek 2011-6-21
I was told to treat the majority of the code as a black box. the functions that determine beam1 and beam2 are,
Psb(N-x)=Psb(N-x+1)+gg1;
Psf(N-x)=Psf(N-x+1)+g1;
Psf(1) is beam1 in my output, and Psb(1) is beam2.
How would I go about using those functions with fmincon? Everything I have found online describing fmincon is very confusing. Thanks again!
Andrew Newell
Andrew Newell 2011-6-21
By itself your code doesn't determine beam1 or beam2. It is a recursive definition for Psb and Psf. Normally such definitions start with f(i) = f0 for i=1 and then work upwards to (say) i=N. You seem to be working downwards from some value. What is that value?
Walter Roberson
Walter Roberson 2011-6-21
If those are definitions for Psb and Psf, then Psb and Psf are linear. You can for example rewrite as
Psb(N-x+1) = Psb(N-x)-gg1
which makes it clear that Psb decreases by gg1 for each increase of 1 in its argument, and so Psb(s) = c(N) - gg1 * x for some constant c dependent upon N.
Derek
Derek 2011-6-21
I am working downwards from N with Psb(N)=Psf(N)=100e-6
Andrew Newell
Andrew Newell 2011-6-21
So what are you trying to vary to optimize |beam1-beam2|?
Walter Roberson
Walter Roberson 2011-6-21
Are gg1 and g1 known constants? Are they variables to be determined but which are not not dependent on N or x ? Or are they a series of values indexed by (N-x) ? Or ... ?
Andrew Newell
Andrew Newell 2011-6-21
To sum up the questions, you need a function that inputs x, or whatever your variable is, and outputs |beam1-beam2|.
Derek
Derek 2011-6-21
Yes that is correct. I need to find a "length" value which causes |Psf(1)-Psb(1)|<0.0001.
Currently I am looping through different length values,
for length=1.7:0.01:1.9
and have an if statement telling the program to output the proper length that satisfys the condition as well as outputting Psb(1) and Psf(1).
My loop is outputting length=1.86 (which i already know is correct), I just need a faster way to do this.
Also, I keep getting the error,
"??? Attempted to access Pb1(1880); index must be a positive integer or logical.
Error in ==> cutandpaste2 at 143
Pb1(N)=5.6e-3;"
when I run the loop. It is not always at Pb1(1880) and changes depending on where I begin the loop. I have tried doing a long format (no help). I still get the optimal length with this error but no output for Psb and Psf.
finally for Walter,
g1=nSf*h*(Sesf*n2(N-x+1)-Sasf*n1(N-x+1)-alphaSf/(nSf*Nt))*Psf(N-x+1)*Nt;
gg1=nSb*h*(Sesb*n2(N-x+1)-Sasb*n1(N-x+1)-alphaSb/(nSb*Nt))*Psb(N-x+1)*Nt;
Walter Roberson
Walter Roberson 2011-6-21
In Pb1(N)=5.6e-3; your N has been calculated as an expression involving non-integers. The calculated value is very close to 1880 (or whatever) but is not *exactly* an integer. You should avoid such calculations. Meanwhile, you can work-around using round()

请先登录，再进行评论。

更多回答（1 个）

Andrew Newell
Andrew Newell 2011-6-17

0 个投票

Your problem sounds like an optimization problem. Using the Optimization Toolbox to find the length that minimizes beam1-beam2 would probably be the best way to speed up your code.

3 个评论
显示 1更早的评论 隐藏 1更早的评论

Derek
Derek 2011-6-21
Sorry for the delay, it was the weekend and I just got back to the problem today. I am unable to use the optimization toolbox because I am student. I am looking into other ways of aquiring it.
Walter Roberson
Walter Roberson 2011-6-21
The Student Edition includes the Optimization Toolbox.
http://www.mathworks.com/academia/student_version/details.html
Derek
Derek 2011-6-21
whoops didn't notice that

请先登录，再进行评论。

类别

在 帮助中心 和 File Exchange 中查找有关 Loops and Conditional Statements 的更多信息

产品

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by